Camel cxf: how to marshal incoming message to XML?

Question

I want to serialize the incoming message to XML. I'm starting from the camel-example-cxf-osgi example.

My Route:

JaxbDataFormat jaxb = new JaxbDataFormat();

from("cxf:bean:reportIncident")
 .convertBodyTo(InputReportIncident.class)
 .marshal(jaxb)
 .bean(new MyBean2())
 .transform(constant(ok));

But I'm getting error and I'm at a loss:

java.io.IOException: javax.xml.bind.JAXBException: class org.apache.camel.example.reportincident.InputReportIncident nor any of its super class is  known to this context.

Appreciate any help. Thx.

Update: the "problem" got mysteriously solved. Now my object is converted to XML automatically. I didn't mention it before but I was getting a conversion exception when applying steps that would automatically transform it to a string. Now I'm getting a nice XML w/o having do do anything. My apologies, I've no idea what I did, I'm still relatively new to Camel. — Ya., May 17 '14 at 01:50

score 2 · Answer 1 · edited May 23 '17 at 12:29

2

Alternatively to the solution of Daniel, you could use

final JAXBContext jaxbContext = JAXBContext.newInstance(InputReportIncident.class);
final DataFormat jaxb = new JaxbDataFormat(jaxbContext);

This comes in handy if there is no ObjectFactory class. See Jaxb: How do I generate ObjectFactory class? for more general information about this subject.

edited May 23 '17 at 12:29

Community

1
1

answered May 16 '14 at 15:43

Peter Keller

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score 1 · Answer 2 · edited May 16 '14 at 15:42

1

You need to point your to the package where you keep your jaxb classes, something like

DataFormat jaxb = new JaxbDataFormat("com.acme.model");

The exception says that the jaxb context doesn't know how to marshal the class InputReportIncident.class.

edited May 16 '14 at 15:42

Peter Keller

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answered May 16 '14 at 03:21

Daniel

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Camel cxf: how to marshal incoming message to XML?

2 Answers2